Question

A smooth tunnel is dug along the radius of the Earth that ends at the centre. A ball is released from the surface of Earth along the tunnel. If the coefficient of restitution for the collision between the ball and the surface of the tunnel is $0.2$ then the distance travelled by the ball before the second collision at the centre is $\frac{n R}{S}$ . Find the value of $n$ .

Answer 1

$a= \, -\frac{G M}{R^{3}}x$
$a= \, -\omega ^{2}x$
$v_{c}=\omega A= \, \sqrt{\frac{G M}{R^{3}}}\times R$

After collision velocity of the ball towards A,
$v_{C}=ev_{C}=0.2\sqrt{\frac{G M}{R}}=\frac{1}{5}\sqrt{\frac{G M}{R}}$
Let now amplitude be A, then
$A^{′}=\frac{v_{C}}{\omega }=\frac{\frac{1}{5} \sqrt{\frac{G M}{R}}}{\sqrt{\frac{G M}{R^{3}}}}=\frac{R}{5}$
Net distance $=R+\left(\frac{R}{5}\right)+\left(\frac{R}{5}\right)=\frac{7 R}{5}$